Question: $ A = \left[\begin{array}{rr}-1 & 5 \\ 3 & -2\end{array}\right]$ $ E = \left[\begin{array}{rrr}1 & -1 & 3 \\ -2 & 1 & 4\end{array}\right]$ What is $ A E$ ?
Answer: Because $ A$ has dimensions $(2\times2)$ and $ E$ has dimensions $(2\times3)$ , the answer matrix will have dimensions $(2\times3)$ $ A E = \left[\begin{array}{rr}{-1} & {5} \\ {3} & {-2}\end{array}\right] \left[\begin{array}{rrr}{1} & \color{#DF0030}{-1} & \color{#9D38BD}{3} \\ {-2} & \color{#DF0030}{1} & \color{#9D38BD}{4}\end{array}\right] = \left[\begin{array}{rrr}? & ? & ? \\ ? & ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ A$ , with the corresponding elements in column $j$ of the second matrix, $ E$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ A$ with the first element in ${\text{column }1}$ of $ E$ , then multiply the second element in ${\text{row }1}$ of $ A$ with the second element in ${\text{column }1}$ of $ E$ , and so on. Add the products together. $ \left[\begin{array}{rrr}{-1}\cdot{1}+{5}\cdot{-2} & ? & ? \\ ? & ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ A$ with the corresponding elements in ${\text{column }1}$ of $ E$ and add the products together. $ \left[\begin{array}{rrr}{-1}\cdot{1}+{5}\cdot{-2} & ? & ? \\ {3}\cdot{1}+{-2}\cdot{-2} & ? & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ A$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ E$ and add the products together. $ \left[\begin{array}{rrr}{-1}\cdot{1}+{5}\cdot{-2} & {-1}\cdot\color{#DF0030}{-1}+{5}\cdot\color{#DF0030}{1} & ? \\ {3}\cdot{1}+{-2}\cdot{-2} & ? & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rrr}{-1}\cdot{1}+{5}\cdot{-2} & {-1}\cdot\color{#DF0030}{-1}+{5}\cdot\color{#DF0030}{1} & {-1}\cdot\color{#9D38BD}{3}+{5}\cdot\color{#9D38BD}{4} \\ {3}\cdot{1}+{-2}\cdot{-2} & {3}\cdot\color{#DF0030}{-1}+{-2}\cdot\color{#DF0030}{1} & {3}\cdot\color{#9D38BD}{3}+{-2}\cdot\color{#9D38BD}{4}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rrr}-11 & 6 & 17 \\ 7 & -5 & 1\end{array}\right] $